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Question Number 185880 by cortano1 last updated on 29/Jan/23

$$\underset{0}{\stackrel{\pi /2}{\int }}\frac{\sqrt[3]{\tan x}}{{(\sin x+\cos x)}^2}dx=?$$

Commented by MJS_new last updated on 29/Jan/23

$$\mathrm{simply}\mathrm{use}t=\sqrt[3]{\tan x}$$

Commented by cortano1 last updated on 29/Jan/23

Answered by cortano1 last updated on 29/Jan/23

$$I=\int \frac{\sqrt[3]{\tan x}}{{(\sin x+\cos x)}^2}dx$$$$=\int \frac{\sqrt[3]{\tan x}\sec {}^{2}x}{{(\tan x+1)}^2}dx$$$$=\int \frac{u^{1/3}}{{(u+1)}^2}du$$

Commented by cortano1 last updated on 29/Jan/23

$${it}^{\prime }scorrect?$$

Answered by MJS_new last updated on 29/Jan/23

$$\underset{0}{\stackrel{\pi /2}{\int }}\frac{\sqrt[3]{\tan x}}{(\sin x+\cos x)}dx=$$$$[t=\sqrt[3]{\tan x}\to dx={3\cos }^{2}x\sqrt[3]{{\tan }^{2}x}dx]$$$$=3\underset{0}{\stackrel{\mathrm{\infty }}{\int }}\frac{t^3}{{(t^3+1)}^2}dt=$$$$\left[{\mathrm{Ostrogradski}}^{\prime }\mathrm{s}\mathrm{Method}\right]$$$$={[-\frac{t}{t^3+1}]}_0^{\mathrm{\infty }}+\underset{0}{\stackrel{\mathrm{\infty }}{\int }}\frac{dt}{t^3+1}=$$$$=0+{[\frac{1}{6}\ln \frac{{(t+1)}^2}{t^2-t+1}+\frac{\sqrt{3}}{3}\arctan \frac{\sqrt{3}(2t-1)}{3}]}_0^{\mathrm{\infty }}=$$$$=\frac{2\sqrt{3}}{9}\pi$$

Commented by cortano1 last updated on 30/Jan/23

$$thankyou$$

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